If a buffer solution is 0.290 M in a weak base (#K_b = 6.4 xx 10^(-5)#) and 0.520 M in its conjugate acid, what is the pH?

Question

Naomi Aronson · Accepted Answer

#"pH" = 9.56#

The first thing to do here is use the base dissociation constant, #K_b#, to determine the #pK_b# of the weak base.

You ought to be aware of that.

#color(purple)(|bar(ul(color(white)(a/a)color(black)(pK_b = - log(K_b))color(white)(a/a)|)))#

Plug in the given #K_b# to get

#pK_b = -log(6.4 * 10^(-5)) = 4.19#

Thus, your buffer solution has similar concentrations of a weak base and its conjugate acid. You are aware that the Henderson - Hasselbalch equation can be used to determine the pOH of a buffer solution containing a weak abase and its conjugate acid.

#color(blue)(|bar(ul(color(white)(a/a)"pOH" = pK_b + log( (["conjugate acid"])/(["weak base"]))color(white)(a/a)|)))#

Notice that when you have equal amounts of weak base and conjugate acid, the pOH of the solution is equal to #pK_b#.

In your case, you have more conjugate acid than weak base, so right from the start you should expect the pOH of the solution to be higher than #pK_b#.

The presence of more conjugate acid implies that the solution is more acidic, i.e. the pH is lower than what corresponds to #"pOH" = pK_b#.

Enter your values now to determine the solution's pOH.

#"pOH" = pK_b + log( (0.520 color(red)(cancel(color(black)("M"))))/(0.290color(red)(cancel(color(black)("M")))))#

#"pOH" = 4.19 + log(0.520/0.290) = 4.44#

Given that a room-temperature aqueous solution has

#color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))#

One could say that the buffer's pH is

#"pH" = 14 - "pOH"#

#"pH" = 14 - 4.44 = color(green)(|bar(ul(color(white)(a/a)9.56color(white)(a/a)|)))#

Jose Ayala · Answer

undefined To find the pH of the buffer solution, use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Given:
Kb = 6.4 × 10^(-5) (Kb = Kw/Ka, where Kw = 1.0 × 10^(-14) at 25°C)
[A-] = 0.290 M (concentration of weak base)
[HA] = 0.520 M (concentration of conjugate acid)

Calculate Ka:
Ka = Kw/Kb = 1.0 × 10^(-14) / 6.4 × 10^(-5)

Calculate pKa:
pKa = -log(Ka)

Calculate pH:
pH = pKa + log([A-]/[HA])